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m11#9

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m11#9 [#permalink] New post 01 Nov 2008, 07:23
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64% (02:03) correct 36% (00:35) wrong based on 14 sessions
Which of the following sets must have the same standard deviation as set {a, b, c}?

A. {ab, b^2, cb}
B. {2a, b + a, c + b}
C. {0, b + a, c - a}
D. {ab, bc, ac}
E. {ab + c, a(1 + b), b(1+a)}

(C) 2008 GMAT Club - m11#9

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[Reveal] Spoiler: OA

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Re: m11#9 [#permalink] New post 01 Nov 2008, 10:51
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The standard deviation of a set does not change if a constant is added to all the members.

Thus, standard deviation of (a,b,c) will be the same as of (a+ab, b+ab, c+ab).

And, option E is the same as (a+ab, b+ab, c+ab).
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Re: m11#9 [#permalink] New post 26 Dec 2012, 05:12
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amitdgr wrote:
Which of the following sets has the same standard deviation as set (a, b, c)?

(C) 2008 GMAT Club - m11#9

* (ab, b^2, cb)
* (2a, b + a, c + b)
* (0, b + a, c - a)
* (ab, bc, ac)
* (ab + c, a(1 + b), b(1+a))

[Reveal] Spoiler: OA
E

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http://gmatclub.com/tests/m11#expl9

I think the explanation is missing/incomplete. Please help me through this problem.


Which of the following sets must have the same standard deviation as set {a, b, c}?

A. {ab, b^2, cb}
B. {2a, b + a, c + b}
C. {0, b + a, c - a}
D. {ab, bc, ac}
E. {ab + c, a(1 + b), b(1+a)}

If we add or subtract a constant to each term in a set the standard deviation will not change.

Notice that set {(ab + c, a(1 + b), b(1+a)}={c+ab, a+ab, b+ab}, so this set is obtained by adding some number ab to each term of set {a, b, c}, which means that those sets must have the same standard deviation.

Answer: E.
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Re: m11#9 [#permalink] New post 28 Oct 2013, 06:22
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Awesome and crisp approach by scthakur and Bunuel..Great work
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Re: m11#9 [#permalink] New post 01 Nov 2008, 11:53
scthakur wrote:
The standard deviation of a set does not change if a constant is added to all the members.

Thus, standard deviation of (a,b,c) will be the same as of (a+ab, b+ab, c+ab).

And, option E is the same as (a+ab, b+ab, c+ab).



Beautiful approach by scthakur. Thats the best approach to this question. +1.

SD of a, b and c and (a+x), (b+x) and (c + x) is the same.
Trying to find exactly what is the SD of a, b and c and the same of each of the options in the question doesnot help solve this question. What helps is understanding the question.
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Re: m11#9 [#permalink] New post 22 Dec 2010, 06:02
My approach was actually using real numbers such as a=2, b=3 and c=4. Though abit lengthy it worked since I applied the rule, the less spread out my answers were the closer my answer was making it E. Thanx now I know another rule;The standard deviation of a set does not change if a constant is added to all the members.
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Re: m11#9 [#permalink] New post 29 Dec 2010, 05:38
E, addition or subtraction of Same constant term does not change standard deviation of the numbers
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Re: m11#9 [#permalink] New post 27 Dec 2011, 21:07
Standard Deviation is the spread of numbers. question is asking which spread of letters equals a, b, c.

I picked numbers 2, 4, 6 for a, b, c.
Plugged in to find another set that has the same SD of 2. E is the only one that worked.

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Re: m11#9 [#permalink] New post 19 Jan 2013, 06:41
Plugging numbers in it's not so time wasting, even though it is prone to errors.

I put a=1, b=2, c=3 with a S.D of +/- 1

A = 2,4,6
B = 2,3,5
C = 0,3,2
D = 2,6,3
E = 5,3,4

E is the only set that has its numbers spread one integer apart.
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Re: m11#9 [#permalink] New post 20 Dec 2013, 18:33
I took a similar, although longer, approach to solving this problem as the person above me. Immediately understanding that this problem was evaluating the spread, I relied on the use of "plugging" numbers in for a,b,c (1,2,3) and then looked for a similar spread amongst the answer choices.

Having read and followed the Manhattan Advanced Quant books, I first started with E and realized that this is the right answer --> matches to my "target"

Would of been even quicker if I would of realized that "ab" is consistent, a constant, throughout the 3 terms; and adding a constant to the terms does not alter the spread. Thanks for the clarification on this one guys!
Re: m11#9   [#permalink] 20 Dec 2013, 18:33
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